Correct Answer: C Let the area of $\triangle AOB$ be $S_1 = 9 \text{ cm}^2$ and the area of $\triangle COD$ be $S_2 = 16 \text{ cm}^2$. Step 1: Identify similar triangles. Since $AB \parallel CD$, we have: $\angle OAB = \angle OCD$ (Alternate interior angles) $\angle OBA = \angle ODC$ (Alternate interior angles) $\angle AOB = \angle COD$ (Vertically opposite angles) Therefore, $\triangle AOB \sim \triangle COD$ by AAA similarity criterion. Step 2: Use the property of areas of similar triangles. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. $$\frac{\text{Area}(\triangle AOB)}{\text{Area}(\triangle COD)} = \left(\frac{AO}{CO}\right)^2 = \left(\frac{BO}{DO}\right)^2 = \left(\frac{AB}{CD}\right)^2$$ Substituting the given values: $$\frac{9}{16} = \left(\frac{AO}{CO}\right)^2$$ Taking the square root of both sides: $$\frac{AO}{CO} = \frac{3}{4}$$ Similarly, $\frac{BO}{DO} = \frac{3}{4}$. Step 3: Relate the areas of adjacent triangles. Consider $\triangle AOD$ and $\triangle COD$. These two triangles share the same height from vertex $D$ to the base $AC$. Therefore, the ratio of their areas is equal to the ratio of their bases on the same line $AC$. $$\frac{\text{Area}(\triangle AOD)}{\text{Area}(\triangle COD)} = \frac{AO}{CO}$$ We found $\frac{AO}{CO} = \frac{3}{4}$, so: $$\frac{\text{Area}(\triangle AOD)}{16} = \frac{3}{4}$$ $$\text{Area}(\triangle AOD) = \frac{3}{4} \times 16 = 12 \text{ cm}^2$$ Step 4: Use the property that areas of triangles on the same base between parallel lines are equal. In a trapezium $ABCD$ with $AB \parallel CD$, the triangles $\triangle ADC$ and $\triangle BDC$ have the same base $CD$ and lie between the parallel lines $AB$ and $CD$. Thus, their heights are equal, and $\text{Area}(\triangle ADC) = \text{Area}(\triangle BDC)$. We can write: $\text{Area}(\triangle ADC) = \text{Area}(\triangle AOD) + \text{Area}(\triangle COD)$ $\text{Area}(\triangle BDC) = \text{Area}(\triangle BOC) + \text{Area}(\triangle COD)$ Since $\text{Area}(\triangle ADC) = \text{Area}(\triangle BDC)$, it follows that: $\text{Area}(\triangle AOD) + \text{Area}(\triangle COD) = \text{Area}(\triangle BOC) + \text{Area}(\triangle COD)$ Which implies: $\text{Area}(\triangle AOD) = \text{Area}(\triangle BOC)$ So, $\text{Area}(\triangle BOC) = 12 \text{ cm}^2$. Alternative derivation for $\text{Area}(\triangle AOD)$ and $\text{Area}(\triangle BOC)$: Let $X = \text{Area}(\triangle AOD) = \text{Area}(\triangle BOC)$. From Step 2, we have $\frac{AO}{CO} = \frac{BO}{DO} = \frac{3}{4}$. Consider $\triangle AOB$ and $\triangle AOD$. They share the same height from vertex $A$ to the line $BD$. $$\frac{\text{Area}(\triangle AOB)}{\text{Area}(\triangle AOD)} = \frac{BO}{DO} = \frac{3}{4}$$ $$\frac{9}{X} = \frac{3}{4} \implies X = \frac{9 \times 4}{3} = 12 \text{ cm}^2$$ Similarly, consider $\triangle BOC$ and $\triangle COD$. They share the same height from vertex $C$ to the line $BD$. $$\frac{\text{Area}(\triangle BOC)}{\text{Area}(\triangle COD)} = \frac{BO}{DO} = \frac{3}{4}$$ $$\frac{X}{16} = \frac{3}{4} \implies X = \frac{16 \times 3}{4} = 12 \text{ cm}^2$$ Both methods yield $X = 12 \text{ cm}^2$. Also, it is a known property that for a trapezium, $X = \sqrt{S_1 S_2} = \sqrt{9 \times 16} = \sqrt{144} = 12 \text{ cm}^2$. Step 5: Calculate the total area of the trapezium. The area of the trapezium $ABCD$ is the sum of the areas of the four triangles it comprises: $$\text{Area}(ABCD) = \text{Area}(\triangle AOB) + \text{Area}(\triangle BOC) + \text{Area}(\triangle COD) + \text{Area}(\triangle AOD)$$ $$\text{Area}(ABCD) = 9 + 12 + 16 + 12$$ $$\text{Area}(ABCD) = 49 \text{ cm}^2$$ This can also be expressed by the formula: $\text{Area}(ABCD) = (\sqrt{S_1} + \sqrt{S_2})^2$. $\text{Area}(ABCD) = (\sqrt{9} + \sqrt{16})^2 = (3 + 4)^2 = 7^2 = 49 \text{ cm}^2$. The final answer is $\boxed{49 \text{ cm}^2}$.